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The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF.
Similarly the largest equilateral triangle which fits into a circle is LMN and PQR is an equilateral triangle with P and Q on the line LM and R on the circumference of the circle. Show that LM = 3PQ

Two ladders are propped up against facing walls. The end of the
first ladder is 10 metres above the foot of the first wall. The end
of the second ladder is 5 metres above the foot of the second wall.
At what height do the ladders cross?

Napkin

Stage: 4 Challenge Level:

This solution is a combination of the work of
David from Trinity School and several others.

A square is folded so that the corner $E$ coincides with the
midpoint of an opposite edge as shown in the diagram. The length of
the edge of the square is 1 unit and the problem is to find the
lengths and areas of the three triangles.

The table shows the solutions:

Triangle

Lengths of sides
3 : 4 : 5

Ratio of lengths
in 3 triangles

Area

Ratio of areas
of 3 triangles

EGF

3/8 : 1/2 : 5/8

3

3/32

9

CED

1/2 : 2/3 : 5/6

4

1/6

16

CAB

1/8 : 1/6 : 5/24

1

1/96

1

This is David's method:

The edge, $EF = 1/2$. Taking $FG = a$ then $GE = 1-a$

As it has a corner of the paper as part of it, we know $EGF$ is
a right angled triangle, and so $GE^2 = EF^2 + FG^2$, which gives
\[ (1-a)^2 = 1/4 + a^2 \]

so \[a = 3/8.\]

Now that we have the length of $a$, we have the lengths of all
the sides of this triangle: $EF = 1/2$ , $FG = 3/8$ and $GE =
5/8.$

At this point, David uses Trigonometry
to complete the problem, but other solutions just used similar
triangles.

The area of triangle $EFG = 1/2$ base $\times$ height $ =
3/32$.

Using this information and trigonometry, we can work out the
angle $GEF$: $\angle GEF = \tan^{-1}(0.375 / 0.5) = \tan^{-1}0.75 =
36.9^{\circ}.$

taking the angle to the nearest tenth of a degree. Knowing this,
we can work out the angle $DEC=53.1^{\circ}$ and use it with the
length $DE$ to work out the length $DC=2/3.$ The last part needed
of this triangle is $CE$ and, using Pythagoras' theorem, we get $CE
=5/6.$

The last triangle needed is $ABC$. We know the length $BC = 1 -
CE = 1/6.$ We can also work out the angle $BCA $ which is equal to
angle $DCE$ and to angle $FEG.$ We can now use trigonometry to work
out the length $AB = 1/6 \tan BCA = 1/8.$

And so once again, using Pythagoras, we can work out the length
of line $AC$ which is $5/24.$

And so we have the final set of lengths: $AB = 1/8$, $BC = 1/ 6$
and $AC = 5/24.$

The area of triangle $ABC = 1/2 \times 1/8 \times 1/6 =
1/96$.

This means the total area of all the paper with a single
thickness = $26/96 = .270833..$

This can be backed up by working out the area of the trapezium
$ABEG$, subtracting the area of triangle $ABC$ and then multiplying
the result by 2 to give the area of the original square that is now
double thickness.

The total area is $2 \times 35/96 + 26/96= 1$ and so all areas
are worked out and recorded, and the total area of single thickness
paper and double thickness areas are recorded.

Here is the Similar Triangles
method:

Triangles EFG, CDE and ABC are similar triangles. This is how
I know: firstly all three are right-angled triangles. Angle GEF and
EGF add up to 90, as do angles CED and GEF. Therefore angles EGF
and CED must be the same.This tells me that angle DCE must be the
same as GEF, because triangle CDE is right-angled. Angle BCA and
DCE are opposite angles so they must be equal, meaning that angle
BAC is equal to CED. All three triangles have the same three
angles, therefore they are similar.

Triangle EFG has sides in the ratio $3:4:5$, so triangle CDE must
also have sides in the ratio $3:4:5$.

$CD = {1 \over 2} \div 3 \times 4$

$CD = {2 \over 3}$

$CE = {1 \over 2} \div 4 \times 5$

$CE = {5 \over 8}$

$EB$ is a side of the square, so $EB$ is 1 unit.
$BC = 1- {5 \over 6}$
$BC = {1 \over 6}$

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